Mass and Moles of a Substance Chemistry requires a method for determining the numbers of molecules in a given mass of a substance. This allows the chemist to carry out “recipes” for compounds based on the relative numbers of atoms involved. The calculation involving the quantities of reactants and products in a chemical equation is called stoichiometry. 2

Chemical Stoichiometry Stoichiometry – The study of quantities of materials consumed and produced in chemical reactions. Copyright © Cengage Learning. All rights reserved

Copyright © Cengage Learning. All rights reserved Objects behave as though they were all identical. Atoms are too small to count. Need average mass of the object. Copyright © Cengage Learning. All rights reserved

Copyright © Cengage Learning. All rights reserved EXERCISE! A pile of marbles weigh 394.80 g. 10 marbles weigh 37.60 g. How many marbles are in the pile? Average mass of one marble = 37.60 g / 10 marbles = 3.76 g / marble 394.80 g / 3.76 g = 105 marbles Copyright © Cengage Learning. All rights reserved

Atomic mass is the mass of an atom in atomic mass units (amu) Micro World atoms & molecules Macro World grams Atomic mass is the mass of an atom in atomic mass units (amu) By definition: 1 atom 12C “weighs” 12 amu On this scale 1H = 1.008 amu 16O = 16.00 amu

Copyright © Cengage Learning. All rights reserved 12C is the standard for atomic mass, with a mass of exactly 12 atomic mass units (amu). The masses of all other atoms are given relative to this standard. Elements occur in nature as mixtures of isotopes. Carbon = 98.89% 12C 1.11% 13C < 0.01% 14C Copyright © Cengage Learning. All rights reserved

Pg 78b

Average Atomic Mass for Carbon 98.89% of 12 amu + 1.11% of 13.0034 amu = exact number (0.9889)(12 u) + (0.0111)(13.0034 u) = 12.01 amu Copyright © Cengage Learning. All rights reserved

Average Atomic Mass for Carbon Even though natural carbon does not contain a single atom with mass 12.01, for stoichiometric purposes, we can consider carbon to be composed of only one type of atom with a mass of 12.01. This enables us to count atoms of natural carbon by weighing a sample of carbon. Copyright © Cengage Learning. All rights reserved

Schematic Diagram of a Mass Spectrometer Copyright © Cengage Learning. All rights reserved

Heavy Light

Isotopes of the same element have identical chemical properties. Some isotopes are radioactive. Find chlorine on the periodic table. What is the atomic number? 17 What is the mass given? 35.45 This is not the mass number of an isotope.

What is this number? It is called the atomic mass - the weighed average of the masses of the isotopes that make up chlorine. Chlorine consists of chlorine-35 and chlorine-37 in a 3:1 ratio. The weighted average is an average corrected by the relative amounts of each isotope present in nature.

Calculate the atomic mass of naturally occurring chlorine if 75 Calculate the atomic mass of naturally occurring chlorine if 75.77% of chlorine atoms are chlorine-35 and 24.23% of chlorine atoms are chlorine-37. Step 1: Convert the percentage to a decimal fraction. 0.7577 chlorine-35 0.2423 chlorine-37

Step 2: Multiply the decimal fraction by the mass of that isotope to obtain the isotope contribution to the atomic mass. For chlorine-35: 0.7577 x 35.00 amu = 26.52 amu For chlorine-37 0.2423 x 37.00 amu = 8.965 amu Step 3: sum to get the weighted average atomic mass of chlorine = 26.52 amu + 8.965 amu = 35.49 amu

Atomic Weights Calculate the atomic weight of boron, B, from the following data: ISOTOPE ISOTOPIC MASS (amu) FRACTIONAL ABUNDANCE B-10 10.013 0.1978 B-11 11.009 0.8022

Atomic Weights Calculate the atomic weight of boron, B, from the following data: ISOTOPE ISOTOPIC MASS (amu) FRACTIONAL ABUNDANCE B-10 10.013 0.1978 B-11 11.009 0.8022 B-10: 10.013 x 0.1978 = 1.9805 B-11: 11.009 x 0.8022 = 8.8314 10.8119 = 10.812 amu ( = atomic wt.)

Calculate the atomic mass Nitrogen consists of two naturally occurring isotopes. 99.63% nitrogen-14 with a mass of 14.003 amu and 0.37% nitrogen-15 with a mass of 15.000 amu. What is the atomic mass of nitrogen?

Copyright © Cengage Learning. All rights reserved EXERCISE! An element consists of 62.60% of an isotope with mass 186.956 umu and 37.40% of an isotope with mass 184.953 amu. Calculate the average atomic mass and identify the element. 186.2 amu Rhenium (Re) Average Atomic Mass = (0.6260)(186.956 u) + (0.3740)(184.953 u) = 186.2 u The element is rhenium (Re). Copyright © Cengage Learning. All rights reserved

Average atomic mass (6.941)

Mass and Moles of a Substance The Mole Concept A mole is defined as the quantity of a given substance that contains as many molecules or formula units as the number of atoms in exactly 12 grams of carbon–12. The number of atoms in a 12-gram sample of carbon–12 is called Avogadro’s number (to which we give the symbol Na). The value of Avogadro’s number is 6.02 x 1023. 2

Dozen = 12 Pair = 2 The mole (mol) is the amount of a substance that contains as many elementary entities as there are atoms in exactly 12.00 grams of 12C 1 mol = NA = 6.0221367 x 1023 Avogadro’s number (NA)

One Mole of: S C Hg Cu Fe

Copyright © Cengage Learning. All rights reserved The number equal to the number of carbon atoms in exactly 12 grams of pure 12C. 1 mole of something consists of 6.022 × 1023 units of that substance (Avogadro’s number). 1 mole C = 6.022 × 1023 C atoms = 12.01 g C Copyright © Cengage Learning. All rights reserved

Copyright © Cengage Learning. All rights reserved EXERCISE! Calculate the number of iron atoms in a 4.48 mole sample of iron. 2.70×1024 Fe atoms (4.48 mol Fe) × (6.022×1023 Fe atoms / 1 mol Fe) = 2.70×1024 Fe atoms. Copyright © Cengage Learning. All rights reserved

Molar mass is the mass of 1 mole of in grams marbles atoms eggs shoes Molar mass is the mass of 1 mole of in grams marbles atoms 1 mole 12C atoms = 6.022 x 1023 atoms = 12.00 g 1 12C atom = 12.00 amu 1 mole 12C atoms = 12.00 g 12C 1 mole lithium atoms = 6.941 g of Li For any element atomic mass (amu) = molar mass (grams)

Mass and Moles of a Substance The molar mass of a substance is the mass of one mole of a substance. For all substances, molar mass, in grams per mole, is numerically equal to the formula weight in atomic mass units. That is, one mole of any element weighs its atomic mass in grams. 2

Copyright © Cengage Learning. All rights reserved Mass in grams of one mole of the substance: Molar Mass of N = 14.01 g/mol Molar Mass of H2O = 18.02 g/mol (2 × 1.008 g) + 16.00 g Molar Mass of Ba(NO3)2 = 261.35 g/mol 137.33 g + (2 × 14.01 g) + (6 × 16.00 g) Copyright © Cengage Learning. All rights reserved

1 12C atom 12.00 amu 12.00 g 6.022 x 1023 12C atoms = 1.66 x 10-24 g 1 amu x 1 amu = 1.66 x 10-24 g or 1 g = 6.022 x 1023 amu M = molar mass in g/mol NA = Avogadro’s number

Copyright © Cengage Learning. All rights reserved CONCEPT CHECK! Which of the following is closest to the average mass of one atom of copper? a) 63.55 g b) 52.00 g c) 58.93 g d) 65.38 g e) 1.055 x 10-22 g The correct answer is “e”. The mass of one atom of copper is going to be extremely small, so only letter “e” makes sense. The calculated solution is (1 Cu atom) × (1 mol Cu/6.022×1023 Cu atoms) × (63.55 g Cu/1 mol Cu). Copyright © Cengage Learning. All rights reserved

Mass and Moles of a Substance Mole calculations Suppose we have 100.0 grams of iron (Fe). The atomic weight of iron is 55.8 g/mol. How many moles of iron does this represent? 2

Mass and Moles of a Substance Mole calculations Conversely, suppose we have 5.75 moles of magnesium (atomic wt. = 24.3 g/mol). What is its mass? 2

Do You Understand Molar Mass? How many atoms are in 0.551 g of potassium (K) ? 1 mol K = 39.10 g K 1 mol K = 6.022 x 1023 atoms K 1 mol K 39.10 g K x x 6.022 x 1023 atoms K 1 mol K = 0.551 g K 8.49 x 1021 atoms K

Let's Practice Using these Tools 1. What is the mass in grams of 2.5 mol Na. 2. Calculate the number of atoms in 1.7 mol B. 3. Calculate the number of atoms in 5.0 g Al. 4. Calculate the mass of 5,000,000 atoms of Au.

Molecular Weight and Formula Weight The formula weight of a substance is the sum of the atomic weights of all the atoms in one formula unit of the compound, whether molecular or not. For example, one formula unit of NaCl contains 1 sodium atom (23.0 amu) and one chlorine atom (35.5 amu), giving a formula weight of 58.5 amu. 2

formula mass (amu) = molar mass (grams) Formula mass is the sum of the atomic masses (in amu) in a formula unit of an ionic compound. NaCl 1Na 22.99 amu 1Cl + 35.45 amu NaCl 58.44 amu For any ionic compound formula mass (amu) = molar mass (grams) 1 formula unit NaCl = 58.44 amu 1 mole NaCl = 58.44 g NaCl

Do You Understand Formula Mass? What is the formula mass of Ca3(PO4)2 ? 1 formula unit of Ca3(PO4)2 3 Ca 3 x 40.08 2 P 2 x 30.97 8 O + 8 x 16.00 310.18 amu

molecular mass (amu) = molar mass (grams) Molecular mass (or molecular weight) is the sum of the atomic masses (in amu) in a molecule. SO2 1S 32.07 amu 2O + 2 x 16.00 amu SO2 64.07 amu For any molecule molecular mass (amu) = molar mass (grams) 1 molecule SO2 = 64.07 amu 1 mole SO2 = 64.07 g SO2

Mass and Moles of a Substance Mole calculations This same method applies to compounds. Suppose we have 100.0 grams of H2O (molecular weight = 18.0 g/mol). How many moles does this represent? 2

Mass and Moles of a Substance Mole calculations Conversely, suppose we have 3.25 moles of glucose, C6H12O6 (molecular wt. = 180.0 g/mol). What is its mass? 2

Do You Understand Molecular Mass? How many H atoms are in 72.5 g of C3H8O ? 1 mol C3H8O = (3 x 12) + (8 x 1) + 16 = 60 g C3H8O 1 mol C3H8O molecules = 8 mol H atoms 1 mol H = 6.022 x 1023 atoms H 1 mol C3H8O 60 g C3H8O x 8 mol H atoms 1 mol C3H8O x 6.022 x 1023 H atoms 1 mol H atoms x = 72.5 g C3H8O 5.82 x 1024 atoms H

Copyright © Cengage Learning. All rights reserved CONCEPT CHECK! Calculate the number of copper atoms in a 63.55 g sample of copper. 6.022×1023 Cu atoms 6.022×1023 Cu atoms; 63.55 g of Cu is 1 mole of Cu, which is equal to Avogadro’s number. Copyright © Cengage Learning. All rights reserved

Copyright © Cengage Learning. All rights reserved CONCEPT CHECK! Which of the following 100.0 g samples contains the greatest number of atoms? Magnesium Zinc Silver The correct answer is “a”. Magnesium has the smallest average atomic mass (24.31 u). Since magnesium is lighter than zinc and silver, it will take more atoms to reach 100.0 g. Copyright © Cengage Learning. All rights reserved

Copyright © Cengage Learning. All rights reserved EXERCISE! Rank the following according to number of atoms (greatest to least): 107.9 g of silver 70.0 g of zinc 21.0 g of magnesium b) a) c) b, a, c; The greater the number of moles, the greater the number of atoms present. Zinc contains 1.07 mol, Ag contains 1.000 mol, and Mg contains 0.864 mol. Copyright © Cengage Learning. All rights reserved

Copyright © Cengage Learning. All rights reserved EXERCISE! Consider separate 100.0 gram samples of each of the following:  H2O, N2O, C3H6O2, CO2 Rank them from greatest to least number of oxygen atoms. H2O, CO2, C3H6O2, N2O H2O, CO2, C3H6O2, N2O; The number of oxygen atoms in each is: H2O = 3.343×1024 CO2 = 2.737×1024 C3H6O2 = 1.626×1024 N2O = 1.368×1024 Copyright © Cengage Learning. All rights reserved

Conceptual Problem Solving Where are we going? Read the problem and decide on the final goal. How do we get there? Work backwards from the final goal to decide where to start. Reality check. Does my answer make sense? Is it reasonable? Copyright © Cengage Learning. All rights reserved

Determining Chemical Formulas The percent composition of a compound is the mass percentage of each element in the compound. We define the mass percentage of “A” as the parts of “A” per hundred parts of the total, by mass. That is, 2

Mass Percentages from Formulas Let’s calculate the percent composition of butane, C4H10. First, we need the molecular mass of C4H10. Now, we can calculate the percents. 2

Percent composition of an element in a compound = n x molar mass of element molar mass of compound x 100% n is the number of moles of the element in 1 mole of the compound %C = 2 x (12.01 g) 46.07 g x 100% = 52.14% C2H6O %H = 6 x (1.008 g) 46.07 g x 100% = 13.13% %O = 1 x (16.00 g) 46.07 g x 100% = 34.73% 52.14% + 13.13% + 34.73% = 100.0%

Copyright © Cengage Learning. All rights reserved Mass percent of an element: For iron in iron(III) oxide, (Fe2O3): Copyright © Cengage Learning. All rights reserved

Determining Chemical Formulas Determining the formula of a compound from the percent composition. The percent composition of a compound leads directly to its empirical formula. An empirical formula (or simplest formula) for a compound is the formula of the substance written with the smallest integer (whole number) subscripts. 2

Copyright © Cengage Learning. All rights reserved Formulas Empirical formula = CH Simplest whole-number ratio Molecular formula = (empirical formula)n [n = integer] Molecular formula = C6H6 = (CH)6 Actual formula of the compound Copyright © Cengage Learning. All rights reserved

A molecular formula shows the exact number of atoms of each element in the smallest unit of a substance An empirical formula shows the simplest whole-number ratio of the atoms in a substance H2O molecular empirical H2O C6H12O6 CH2O O3 O N2H4 NH2

Percent Composition and Empirical Formulas Determine the empirical formula of a compound that has the following percent composition by mass: K 24.75, Mn 34.77, O 40.51 percent. nK = 24.75 g K x = 0.6330 mol K 1 mol K 39.10 g K nMn = 34.77 g Mn x = 0.6329 mol Mn 1 mol Mn 54.94 g Mn nO = 40.51 g O x = 2.532 mol O 1 mol O 16.00 g O

Percent Composition and Empirical Formulas nK = 0.6330, nMn = 0.6329, nO = 2.532 K : ~ 1.0 0.6330 0.6329 Mn : 0.6329 = 1.0 O : ~ 4.0 2.532 0.6329 KMnO4

g of O = g of sample – (g of C + g of H) 4.0 g O = 0.25 mol O Combust 11.5 g ethanol Collect 22.0 g CO2 and 13.5 g H2O g CO2 mol CO2 mol C g C 6.0 g C = 0.5 mol C g H2O mol H2O mol H g H 1.5 g H = 1.5 mol H g of O = g of sample – (g of C + g of H) 4.0 g O = 0.25 mol O Empirical formula C0.5H1.5O0.25 Divide by smallest subscript (0.25) Empirical formula C2H6O

Determining Chemical Formulas Determining the empirical formula from the percent composition. Benzoic acid is a white, crystalline powder used as a food preservative. The compound contains 68.8% C, 5.0% H, and 26.2% O by mass. What is its empirical formula? In other words, give the smallest whole-number ratio of the subscripts in the formula Cx HyOz 2

Determining Chemical Formulas Determining the empirical formula from the percent composition. Our 100.0 grams of benzoic acid would contain: This isn’t quite a whole number ratio, but if we divide each number by the smallest of the three, a better ratio might emerge. 2

Determining Chemical Formulas Determining the empirical formula from the percent composition. Our 100.0 grams of benzoic acid would contain: now it’s not too difficult to see that the smallest whole number ratio is 7:6:2. The empirical formula is C7H6O2 . 2

Analyzing for Carbon and Hydrogen Device used to determine the mass percent of each element in a compound. Copyright © Cengage Learning. All rights reserved

Determining Chemical Formulas Determining the “true” molecular formula from the empirical formula. An empirical formula gives only the smallest whole-number ratio of atoms in a formula. The “true” molecular formula could be a multiple of the empirical formula (since both would have the same percent composition). To determine the “true” molecular formula, we must know the “true” molecular weight of the compound. 2

Determining Chemical Formulas Determining the “true” molecular formula from the empirical formula. For example, suppose the empirical formula of a compound is CH2O and its “true” molecular weight is 60.0 g/mol. The molar weight of the empirical formula (the “empirical weight”) is only 30.0 g/mol. This would imply that the “true” molecular formula is actually the empirical formula doubled, or C2H4O2 2

Copyright © Cengage Learning. All rights reserved EXERCISE! The composition of adipic acid is 49.3% C, 6.9% H, and 43.8% O (by mass). The molar mass of the compound is about 146 g/mol. What is the empirical formula? C3H5O2 What is the molecular formula? C6H10O4 Assume 100.0g. 49.3 g of carbon is 4.105 mol C (49.3/12.01). 6.9 g of hydrogen is 6.845 mol H (6.9/1.008). 43.8 g of oxygen is 2.7375 mol O (43.8/16.00). The ratio of C:H:O is 1.5:2.5:1 (C: 4.105/2.7375 = 1.5; H: 6.845/2.7375 = 2.5). Multiplying each by 2 to get a whole number becomes a ratio of 3:5:2. The empirical formula is therefore C3H5O2. The molar mass of the empirical formula is 73.07 g/mol, which goes into the molar mass of the molecular formula 2 times (146/73.07). The molecular formula is therefore C6H10O4. Copyright © Cengage Learning. All rights reserved

Determining Chemical Formulas – Ionic Compounds The percent composition of a compound leads directly to its empirical formula. An empirical formula (or simplest formula) for a ionic compound is the true molecular formula for that compound. 2

3 ways of representing the reaction of H2 with O2 to form H2O A process in which one or more substances is changed into one or more new substances is a chemical reaction A chemical equation uses chemical symbols to show what happens during a chemical reaction 3 ways of representing the reaction of H2 with O2 to form H2O reactants products

How to “Read” Chemical Equations 2 Mg + O2 2 MgO 2 atoms Mg + 1 molecule O2 makes 2 formula units MgO 2 moles Mg + 1 mole O2 makes 2 moles MgO 48.6 grams Mg + 32.0 grams O2 makes 80.6 g MgO IS NOT 2 grams Mg + 1 gram O2 makes 2 g MgO

Features of a Chemical Equation  - means heat is needed. Products and reactants must be specified using chemical symbols Reactants- written on the left of arrow Products - written on the right Physical states are shown in parentheses

All the atoms of every reactant must also appear in the products. Coefficient - how many of the substance are in the reaction The equation must be balanced. All the atoms of every reactant must also appear in the products. How many Hg’s on left? on right? How many O’s on left?

Copyright © Cengage Learning. All rights reserved The balanced equation represents an overall ratio of reactants and products, not what actually “happens” during a reaction. Use the coefficients in the balanced equation to decide the amount of each reactant that is used, and the amount of each product that is formed. Copyright © Cengage Learning. All rights reserved

H2 is released when Na is placed in water. The Experimental Basis of a Chemical Equation We know that a chemical equation represents a chemical change. The following is evidence for a reaction: Release of a gas. CO2 is released when acid is placed in a solution containing CO32- ions. H2 is released when Na is placed in water. Formation of a solid (precipitate.) A solution containing Ag+ ions is mixed with a solution containing Cl- ions.

Acid and base are mixed together Heat is produced or absorbed (temperature changes) Acid and base are mixed together The color changes Light is absorbed or emitted Changes in the way the substances behave in an electrical or magnetic field Changes in electrical properties.

Writing and Balancing the Equation for a Chemical Reaction Determine what reaction is occurring. What are the reactants, the products, and the physical states involved? Write the unbalanced equation that summarizes the reaction described in step 1. Balance the equation by inspection, starting with the most complicated molecule(s). The same number of each type of atom needs to appear on both reactant and product sides. Do NOT change the formulas of any of the reactants or products. Copyright © Cengage Learning. All rights reserved

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Chemical Reactions: Equations Writing chemical equations A chemical equation is the symbolic representation of a chemical reaction in terms of chemical formulas. For example, the burning of sodium and chlorine to produce sodium chloride is written The reactants are starting substances in a chemical reaction. The arrow means “yields.” The formulas on the right side of the arrow represent the products. 2

Chemical Reactions: Equations In many cases, it is useful to indicate the states of the substances in the equation. Writing chemical equations When you use these labels, the previous equation becomes 2

Chemical Reactions: Equations The law of conservation of mass dictates that the total number of atoms of each element on both sides of a chemical equation must match. The equation is then said to be balanced. Writing chemical equations Consider the combustion of methane to produce carbon dioxide and water. 2

Chemical Reactions: Equations Writing chemical equations For this equation to balance, two molecules of oxygen must be consumed for each molecule of methane, producing one molecule of CO2 and two molecules of water. Now the equation is “balanced.” 2 2

Balancing Chemical Equations Write the correct formula(s) for the reactants on the left side and the correct formula(s) for the product(s) on the right side of the equation. Ethane reacts with oxygen to form carbon dioxide and water C2H6 + O2 CO2 + H2O Change the numbers in front of the formulas (coefficients) to make the number of atoms of each element the same on both sides of the equation. Do not change the subscripts. 2C2H6 NOT C4H12

Balancing Chemical Equations Start by balancing those elements that appear in only one reactant and one product. C2H6 + O2 CO2 + H2O start with C or H but not O 1 carbon on right 2 carbon on left multiply CO2 by 2 C2H6 + O2 2CO2 + H2O 6 hydrogen on left 2 hydrogen on right multiply H2O by 3 C2H6 + O2 2CO2 + 3H2O

Balancing Chemical Equations Balance those elements that appear in two or more reactants or products. multiply O2 by 7 2 C2H6 + O2 2CO2 + 3H2O 2 oxygen on left 4 oxygen (2x2) + 3 oxygen (3x1) = 7 oxygen on right C2H6 + O2 2CO2 + 3H2O 7 2 remove fraction multiply both sides by 2 2C2H6 + 7O2 4CO2 + 6H2O

Balancing Chemical Equations Check to make sure that you have the same number of each type of atom on both sides of the equation. 2C2H6 + 7O2 4CO2 + 6H2O 4 C (2 x 2) 4 C 12 H (2 x 6) 12 H (6 x 2) 14 O (7 x 2) 14 O (4 x 2 + 6) Reactants Products 4 C 12 H 14 O

Chemical Reactions: Equations Balance the following equations. 2

Chemical Reactions: Equations Balance the following equations. 2 6 6 2 9 3 4 2

Stoichiometry: Quantitative Relations in Chemical Reactions Stoichiometry is the calculation of the quantities of reactants and products involved in a chemical reaction. It is based on the balanced chemical equation and on the relationship between mass and moles. Such calculations are fundamental to most quantitative work in chemistry. 2

Amounts of Reactants and Products Write balanced chemical equation Convert quantities of known substances into moles Use coefficients in balanced equation to calculate the number of moles of the sought quantity Convert moles of sought quantity into desired units

Molar Interpretation of a Chemical Equation The balanced chemical equation can be interpreted in numbers of molecules, but generally chemists interpret equations as “mole-to-mole” relationships. For example, the Haber process for producing ammonia involves the reaction of hydrogen and nitrogen. 2

Molar Interpretation of a Chemical Equation This balanced chemical equation shows that one mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3. 1 molecule N2 + 3 molecules H2 2 molecules NH3 Because moles can be converted to mass, you can also give a mass interpretation of a chemical equation. 2

Calculations Using the Chemical Equation 7 We will learn in this section to calculate quantities of reactants and products in a chemical reaction. Need a balanced chemical equation for the reaction of interest. Keep in mind that the coefficients represent the number of moles of each substance in the equation.

Molar Interpretation of a Chemical Equation Suppose we wished to determine the number of moles of NH3 we could obtain from 4.8 mol H2. Because the coefficients in the balanced equation represent mole-to-mole ratios, the calculation is simple. 2

Mass Relationships in Chemical Equations Amounts of substances in a chemical reaction by mass. How many grams of HCl are required to react with 5.00 grams manganese dioxide according to this equation? 2

Mass Relationships in Chemical Equations First, you write what is given (5.00 g MnO2) and convert this to moles. Then convert to moles of what is desired.(mol HCl) Finally, you convert this to mass (g HCl) 2

Copyright © Cengage Learning. All rights reserved EXERCISE! Consider the following reaction: If 6.25 g of phosphorus is burned, what mass of oxygen does it combine with? 8.07 g O2 (6.25 g P4)(1 mol P4 / 123.88 g P4)(5 mol O2 / 1 mol P4)(32.00 g O2 / 1 mol O2) = 8.07 g O2 Copyright © Cengage Learning. All rights reserved

Copyright © Cengage Learning. All rights reserved EXERCISE! (Part I) Methane (CH4) reacts with the oxygen in the air to produce carbon dioxide and water. Ammonia (NH3) reacts with the oxygen in the air to produce nitrogen monoxide and water. Write balanced equations for each of these reactions. CH4 + 2O2  CO2 + 2H2O 4NH3 + 5O2  4NO + 6H2O Copyright © Cengage Learning. All rights reserved

Copyright © Cengage Learning. All rights reserved EXERCISE! (Part II) Methane (CH4) reacts with the oxygen in the air to produce carbon dioxide and water. Ammonia (NH3) reacts with the oxygen in the air to produce nitrogen monoxide and water. What mass of ammonia would produce the same amount of water as 1.00 g of methane reacting with excess oxygen? 1.42 g of ammonia is required. Use the balanced equations from the previous exercise to start this problem. Determine the amount of water produced from 1.00 g of methane (which is 0.125 moles of water). Use the moles of water and the balanced equation to determine the amount of ammonia needed. See the “Let’s Think About It” slide next to get the students going. Copyright © Cengage Learning. All rights reserved

Copyright © Cengage Learning. All rights reserved Let’s Think About It Where are we going? To find the mass of ammonia that would produce the same amount of water as 1.00 g of methane reacting with excess oxygen. How do we get there? We need to know: How much water is produced from 1.00 g of methane and excess oxygen. How much ammonia is needed to produce the amount of water calculated above. Copyright © Cengage Learning. All rights reserved

Methanol burns in air according to the equation 2CH3OH + 3O2 2CO2 + 4H2O If 209 g of methanol are used up in the combustion, what mass of water is produced? grams CH3OH moles CH3OH moles H2O grams H2O molar mass CH3OH coefficients chemical equation molar mass H2O 1 mol CH3OH 32.0 g CH3OH x 4 mol H2O 2 mol CH3OH x 18.0 g H2O 1 mol H2O x = 209 g CH3OH 235 g H2O

Let's do some examples. Na + Cl2  NaCl 1. Balance the equation. 2. Calculate the moles of Cl2 which will react with 5.00 mol Na. 3. Calculate the number of grams of NaCl which will be produced when 5.00 mol Na reacts with an excess of Cl2. 4. Calculate the grams of Na which will react with 5.00 g Cl2.

Copyright © Cengage Learning. All rights reserved Limiting Reactants Limiting reactant – the reactant that runs out first and thus limits the amounts of products that can be formed. Determine which reactant is limiting to calculate correctly the amounts of products that will be formed. Copyright © Cengage Learning. All rights reserved

Limiting Reagent The limiting reactant (or limiting reagent) is the reactant that is entirely consumed when the reaction goes to completion. The limiting reagent ultimately determines how much product can be obtained. For example, bicycles require one frame and two wheels. If you have 20 wheels but only 5 frames, it is clear that the number of frames will determine how many bicycles can be made. 2

Limiting Reactants To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE Copyright © Cengage Learning. All rights reserved

A. The Concept of Limiting Reactants Stoichiometric mixture N2(g) + 3H2(g) 2NH3(g)

A. The Concept of Limiting Reactants Limiting reactant mixture N2(g) + 3H2(g)  2NH3(g)

Limiting Reagent Zinc metal reacts with hydrochloric acid by the following reaction. If 0.30 mol Zn is added to hydrochloric acid containing 0.52 mol HCl, how many moles of H2 are produced? 2

Limiting Reagent Take each reactant in turn and ask how much product would be obtained if each were totally consumed. The reactant that gives the smaller amount is the limiting reagent. Since HCl is the limiting reagent, the amount of H2 produced must be 0.26 mol. 2

Do You Understand Limiting Reagents? In one process, 124 g of Al are reacted with 601 g of Fe2O3 2Al + Fe2O3 Al2O3 + 2Fe Calculate the mass of Al2O3 formed. g Al mol Al mol Fe2O3 needed g Fe2O3 needed OR g Fe2O3 mol Fe2O3 mol Al needed g Al needed 1 mol Al 27.0 g Al x 1 mol Fe2O3 2 mol Al x 160. g Fe2O3 1 mol Fe2O3 x = 124 g Al 367 g Fe2O3 Start with 124 g Al need 367 g Fe2O3 Have more Fe2O3 (601 g) so Al is limiting reagent

Use limiting reagent (Al) to calculate amount of product that can be formed. g Al mol Al mol Al2O3 g Al2O3 2Al + Fe2O3 Al2O3 + 2Fe 1 mol Al 27.0 g Al x 1 mol Al2O3 2 mol Al x 102. g Al2O3 1 mol Al2O3 x = 124 g Al 234 g Al2O3

Copyright © Cengage Learning. All rights reserved CONCEPT CHECK! Which of the following reaction mixtures could produce the greatest amount of product? Each involves the reaction symbolized by the equation: 2H2 + O2 2H2O a) 2 moles of H2 and 2 moles of O2 2 moles of H2 and 3 moles of O2 2 moles of H2 and 1 mole of O2 d) 3 moles of H2 and 1 mole of O2 e) Each produce the same amount of product. The correct answer is “e”. All of these reaction mixtures would produce two moles of water. Copyright © Cengage Learning. All rights reserved

Copyright © Cengage Learning. All rights reserved Notice We cannot simply add the total moles of all the reactants to decide which reactant mixture makes the most product. We must always think about how much product can be formed by using what we are given, and the ratio in the balanced equation. Copyright © Cengage Learning. All rights reserved

Copyright © Cengage Learning. All rights reserved CONCEPT CHECK! You know that chemical A reacts with chemical B. You react 10.0 g of A with 10.0 g of B. What information do you need to know in order to determine the mass of product that will be produced? See the “Let’s Think About It” slide next to get the students going. Copyright © Cengage Learning. All rights reserved

Copyright © Cengage Learning. All rights reserved Let’s Think About It Where are we going? To determine the mass of product that will be produced when you react 10.0 g of A with 10.0 g of B. How do we get there? We need to know: The mole ratio between A, B, and the product they form. In other words, we need to know the balanced reaction equation. The molar masses of A, B, and the product they form. Copyright © Cengage Learning. All rights reserved

Copyright © Cengage Learning. All rights reserved EXERCISE! You react 10.0 g of A with 10.0 g of B. What mass of product will be produced given that the molar mass of A is 10.0 g/mol, B is 20.0 g/mol, and C is 25.0 g/mol? They react according to the equation: A + 3B 2C 8.33 g of C is produced. Note: Use the box animations to assist in explaining how to solve the problem. Copyright © Cengage Learning. All rights reserved

Reaction Yield Theoretical Yield is the amount of product that would result if all the limiting reagent reacted. Actual Yield is the amount of product actually obtained from a reaction. % Yield = Actual Yield Theoretical Yield x 100

Theoretical and Percent Yield The theoretical yield of product is the maximum amount of product that can be obtained from given amounts of reactants. The percentage yield is the actual yield (experimentally determined) expressed as a percentage of the theoretical yield (calculated). 2

Theoretical and Percent Yield To illustrate the calculation of percentage yield, recall that the theoretical yield of H2 in the previous example was 0.26 mol (or 0.52 g) H2. If the actual yield of the reaction had been 0.22 g H2, then 2

Copyright © Cengage Learning. All rights reserved EXERCISE! Consider the following reaction: P4(s) + 6F2(g) 4PF3(g) What mass of P4 is needed to produce 85.0 g of PF3 if the reaction has a 64.9% yield? 46.1 g P4 64.9% = (85.0 g PF3 / x)(100); x = 130.97 g PF3 (130.97 g PF3)(1 mol PF3 / 87.97 g PF3)(1 mol P4 / 4 mol PF3)(123.88 g P4 / 1 mol P4) = 46.1 g of P4 needed Copyright © Cengage Learning. All rights reserved

Operational Skills Calculating the formula weight from a formula. Calculating the mass of an atom or molecule. Converting moles of substance to grams and vice versa. Calculating the number of molecules in a given mass. Calculating the percentage composition from the formula. Calculating the mass of an element in a given mass of compound. Calculating the percentages C and H by combustion. Determining the empirical formula from percentage composition. Determining the true molecular formula. Relating quantities in a chemical equation. Calculating with a limiting reagent. 2

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